Simplify and expand the following expression: $ \dfrac{2}{3x - 12}+ \dfrac{2}{5x + 45}+ \dfrac{4}{x^2 + 5x - 36} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{2}{3x - 12} = \dfrac{2}{3(x - 4)}$ We can factor a $5$ out of denominator in the second term: $ \dfrac{2}{5x + 45} = \dfrac{2}{5(x + 9)}$ We can factor the quadratic in the third term: $ \dfrac{4}{x^2 + 5x - 36} = \dfrac{4}{(x - 4)(x + 9)}$ Now we have: $ \dfrac{2}{3(x - 4)}+ \dfrac{2}{5(x + 9)}+ \dfrac{4}{(x - 4)(x + 9)} $ The least common multiple of the denominators is: $ 15(x - 4)(x + 9)$ In order to get the first term over $15(x - 4)(x + 9)$ , multiply by $\dfrac{5(x + 9)}{5(x + 9)}$ $ \dfrac{2}{3(x - 4)} \times \dfrac{5(x + 9)}{5(x + 9)} = \dfrac{10(x + 9)}{15(x - 4)(x + 9)} $ In order to get the second term over $15(x - 4)(x + 9)$ , multiply by $\dfrac{3(x - 4)}{3(x - 4)}$ $ \dfrac{2}{5(x + 9)} \times \dfrac{3(x - 4)}{3(x - 4)} = \dfrac{6(x - 4)}{15(x - 4)(x + 9)} $ In order to get the third term over $15(x - 4)(x + 9)$ , multiply by $\dfrac{15}{15}$ $ \dfrac{4}{(x - 4)(x + 9)} \times \dfrac{15}{15} = \dfrac{60}{15(x - 4)(x + 9)} $ Now we have: $ \dfrac{10(x + 9)}{15(x - 4)(x + 9)} + \dfrac{6(x - 4)}{15(x - 4)(x + 9)} + \dfrac{60}{15(x - 4)(x + 9)} $ $ = \dfrac{ 10(x + 9) + 6(x - 4) + 60} {15(x - 4)(x + 9)} $ Expand: $ = \dfrac{10x + 90 + 6x - 24 + 60}{15x^2 + 75x - 540} $ $ = \dfrac{16x + 126}{15x^2 + 75x - 540}$